3.3.40 \(\int \frac {A+C \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2} \, dx\) [240]

Optimal. Leaf size=170 \[ \frac {4 A \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a^2 d}-\frac {(5 A-C) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{3 a^2 d}-\frac {(5 A-C) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(A+C) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d (a+a \sec (c+d x))^2} \]

[Out]

-1/3*(5*A-C)*sin(d*x+c)*sec(d*x+c)^(1/2)/a^2/d/(1+sec(d*x+c))-1/3*(A+C)*sin(d*x+c)*sec(d*x+c)^(1/2)/d/(a+a*sec
(d*x+c))^2+4*A*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c
)^(1/2)*sec(d*x+c)^(1/2)/a^2/d-1/3*(5*A-C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d
*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a^2/d

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Rubi [A]
time = 0.22, antiderivative size = 170, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {4170, 4105, 3872, 3856, 2719, 2720} \begin {gather*} -\frac {(5 A-C) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 a^2 d (\sec (c+d x)+1)}-\frac {(5 A-C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a^2 d}+\frac {4 A \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d}-\frac {(A+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + C*Sec[c + d*x]^2)/(Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x])^2),x]

[Out]

(4*A*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(a^2*d) - ((5*A - C)*Sqrt[Cos[c + d*x]]*
EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*a^2*d) - ((5*A - C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(3*a^2*d
*(1 + Sec[c + d*x])) - ((A + C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(3*d*(a + a*Sec[c + d*x])^2)

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 3856

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rule 3872

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 4105

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(-(A*b - a*B))*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(b*f*(
2*m + 1))), x] - Dist[1/(a^2*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*Simp[b*B*n - a*A*
(2*m + n + 1) + (A*b - a*B)*(m + n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[
A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0]

Rule 4170

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b
_.) + (a_))^(m_), x_Symbol] :> Simp[(-a)*(A + C)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(a*f*
(2*m + 1))), x] + Dist[1/(a*b*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*Simp[b*C*n + A*b
*(2*m + n + 1) - (a*(A*(m + n + 1) - C*(m - n)))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, C, n}, x
] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]

Rubi steps

\begin {align*} \int \frac {A+C \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2} \, dx &=-\frac {(A+C) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {\int \frac {-\frac {1}{2} a (7 A+C)+\frac {3}{2} a (A-C) \sec (c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))} \, dx}{3 a^2}\\ &=-\frac {(5 A-C) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(A+C) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {\int \frac {-6 a^2 A+\frac {1}{2} a^2 (5 A-C) \sec (c+d x)}{\sqrt {\sec (c+d x)}} \, dx}{3 a^4}\\ &=-\frac {(5 A-C) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(A+C) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac {(2 A) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx}{a^2}-\frac {(5 A-C) \int \sqrt {\sec (c+d x)} \, dx}{6 a^2}\\ &=-\frac {(5 A-C) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(A+C) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac {\left (2 A \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{a^2}-\frac {\left ((5 A-C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{6 a^2}\\ &=\frac {4 A \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a^2 d}-\frac {(5 A-C) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{3 a^2 d}-\frac {(5 A-C) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(A+C) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 3.78, size = 298, normalized size = 1.75 \begin {gather*} \frac {\cos ^4\left (\frac {1}{2} (c+d x)\right ) \left (\frac {8 i e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right ) \left (-C e^{i (c+d x)} \left (-1+e^{i (c+d x)}\right )+A \left (3+16 e^{i (c+d x)}+20 e^{2 i (c+d x)}+9 e^{3 i (c+d x)}\right )\right ) \sqrt {\sec (c+d x)}}{d \left (1+e^{i (c+d x)}\right )^3}-\frac {8 (5 A-C) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{d}-\frac {32 i A e^{i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i (c+d x)}\right ) \sqrt {\sec (c+d x)}}{d}\right ) \left (A+C \sec ^2(c+d x)\right )}{3 a^2 (A+2 C+A \cos (2 (c+d x))) (1+\sec (c+d x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + C*Sec[c + d*x]^2)/(Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x])^2),x]

[Out]

(Cos[(c + d*x)/2]^4*(((8*I)*(1 + E^((2*I)*(c + d*x)))*(-(C*E^(I*(c + d*x))*(-1 + E^(I*(c + d*x)))) + A*(3 + 16
*E^(I*(c + d*x)) + 20*E^((2*I)*(c + d*x)) + 9*E^((3*I)*(c + d*x))))*Sqrt[Sec[c + d*x]])/(d*E^(I*(c + d*x))*(1
+ E^(I*(c + d*x)))^3) - (8*(5*A - C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d - ((32
*I)*A*E^(I*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))]*Sqr
t[Sec[c + d*x]])/d)*(A + C*Sec[c + d*x]^2))/(3*a^2*(A + 2*C + A*Cos[2*(c + d*x)])*(1 + Sec[c + d*x])^2)

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Maple [A]
time = 3.86, size = 352, normalized size = 2.07

method	result	size
default	\(\frac {\sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (24 A \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+10 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\cos ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+24 A \left (\cos ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-2 C \left (\cos ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-38 A \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 C \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+15 A \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3 C \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-A -C \right )}{6 a^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\)	\(352\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2/sec(d*x+c)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/6*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(24*A*cos(1/2*d*x+1/2*c)^6+10*A*(sin(1/2*d*x+1/2*c
)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*cos(1/2*d*x+1/2*c)^3+24*A*c
os(1/2*d*x+1/2*c)^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c
),2^(1/2))-2*C*cos(1/2*d*x+1/2*c)^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticF(c
os(1/2*d*x+1/2*c),2^(1/2))-38*A*cos(1/2*d*x+1/2*c)^4-2*C*cos(1/2*d*x+1/2*c)^4+15*A*cos(1/2*d*x+1/2*c)^2+3*C*co
s(1/2*d*x+1/2*c)^2-A-C)/a^2/cos(1/2*d*x+1/2*c)^3/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*
d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

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Maxima [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2/sec(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.67, size = 326, normalized size = 1.92 \begin {gather*} \frac {{\left (\sqrt {2} {\left (5 i \, A - i \, C\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} {\left (-5 i \, A + i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (5 i \, A - i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + {\left (\sqrt {2} {\left (-5 i \, A + i \, C\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} {\left (5 i \, A - i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-5 i \, A + i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 12 \, {\left (-i \, \sqrt {2} A \cos \left (d x + c\right )^{2} - 2 i \, \sqrt {2} A \cos \left (d x + c\right ) - i \, \sqrt {2} A\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 12 \, {\left (i \, \sqrt {2} A \cos \left (d x + c\right )^{2} + 2 i \, \sqrt {2} A \cos \left (d x + c\right ) + i \, \sqrt {2} A\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {2 \, {\left (6 \, A \cos \left (d x + c\right )^{2} + {\left (5 \, A - C\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2/sec(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

1/6*((sqrt(2)*(5*I*A - I*C)*cos(d*x + c)^2 - 2*sqrt(2)*(-5*I*A + I*C)*cos(d*x + c) + sqrt(2)*(5*I*A - I*C))*we
ierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + (sqrt(2)*(-5*I*A + I*C)*cos(d*x + c)^2 - 2*sqrt(2)*(
5*I*A - I*C)*cos(d*x + c) + sqrt(2)*(-5*I*A + I*C))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))
- 12*(-I*sqrt(2)*A*cos(d*x + c)^2 - 2*I*sqrt(2)*A*cos(d*x + c) - I*sqrt(2)*A)*weierstrassZeta(-4, 0, weierstra
ssPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 12*(I*sqrt(2)*A*cos(d*x + c)^2 + 2*I*sqrt(2)*A*cos(d*x + c
) + I*sqrt(2)*A)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(6*A*co
s(d*x + c)^2 + (5*A - C)*cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*
x + c) + a^2*d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {A}{\sec ^{\frac {5}{2}}{\left (c + d x \right )} + 2 \sec ^{\frac {3}{2}}{\left (c + d x \right )} + \sqrt {\sec {\left (c + d x \right )}}}\, dx + \int \frac {C \sec ^{2}{\left (c + d x \right )}}{\sec ^{\frac {5}{2}}{\left (c + d x \right )} + 2 \sec ^{\frac {3}{2}}{\left (c + d x \right )} + \sqrt {\sec {\left (c + d x \right )}}}\, dx}{a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**2/sec(d*x+c)**(1/2),x)

[Out]

(Integral(A/(sec(c + d*x)**(5/2) + 2*sec(c + d*x)**(3/2) + sqrt(sec(c + d*x))), x) + Integral(C*sec(c + d*x)**
2/(sec(c + d*x)**(5/2) + 2*sec(c + d*x)**(3/2) + sqrt(sec(c + d*x))), x))/a**2

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2/sec(d*x+c)^(1/2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + A)/((a*sec(d*x + c) + a)^2*sqrt(sec(d*x + c))), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {A+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + C/cos(c + d*x)^2)/((a + a/cos(c + d*x))^2*(1/cos(c + d*x))^(1/2)),x)

[Out]

int((A + C/cos(c + d*x)^2)/((a + a/cos(c + d*x))^2*(1/cos(c + d*x))^(1/2)), x)

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